Problem
Show that $V(X) =0$ if and only if there is a constanct $c$ such that $P(X = c) = 1$.
Solution
First, let's note that since we are talking about a discrete $P(X=c)$, $X$ is a discrete r.v. Let's assign these two conditions.
- a: $V(X) = 0$
- b: $P(X=c) = 1$
Part 1: show $b \implies a$
$P(X=c) = 1$ aka $f_X(c) = 1$. Recall that from the basics of probability distributions $\sum_i f_X(i) = 1$ and that every $f_X(i) \geq 0$. If there exists a $c$ such that $f_X(c) = 1$ then every other $f_X(i)$ must be 0.
$E(X) = \mu = f_X(c)c = 1 \times c = c$
$E(X^2) = \sum_i x^2 f_X(x) = c^2 f_x(c) = c^2$
So $V(X) = E(X^2) - \mu^2 = c^2 - c^2 = 0$.
Part 2 (TODO): show $a \implies b$
I got stumped by this only to find that the solution used Chebyshev's inequality, which isn't covered until chapter 4. Screw this!
Sources
- 36-705 hw1 problem 8